Problem: $f(x)= \begin{cases} \dfrac{ x + 3 }{ ( x + 3 )( x - 1 ) } & \text{if } x \neq 1 \\ 4 & \text{if } x = 1 \end{cases}$ What is the domain of the real-valued function $f(x)$ ?
Solution: $f(x)$ is a piecewise function, so we need to examine where each piece is undefined. The first piecewise definition of $f(x)$ $\frac{ x + 3 }{ ( x + 3 )( x - 1 ) }$ , is undefined when its denominator is 0. The denominator is 0 when $x=-3$ or $x=1$ So, based on the first piecewise definition, we know that $x \neq -3$ and $x \neq 1$ However, the second piecewise definition applies when $x = 1$ , and the second piecewise definition, $4$ , has no weird gaps or holes, so $f(x)$ is defined at $x = 1$ So the only restriction on the domain is that $x \neq -3$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \neq-3\, \}$.